Question

vapour pressure of an aqueous solution is 2% less than that of the solvent. The molality of the solution is​

Answer 1

Answer:

the molality of the solution is 1.13m

Explanation:

$\frac{P^{0} -Ps}{P^{2} }$ =P

(taking  pressure of pure solvent as 100 and 98 as solvent as V.P. of solution is 2% less)

$\frac{100-98}{100}$

$\frac{2}{100}$

(The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.)

hence,

Xs =  $\frac{m}{m+\frac{1000}{M.W. of solvent} }$

(as it is a aqueous solution solvent is water)

$\frac{2}{100}$ = $\frac{m}{m+\frac{1000}{18} }$

$\frac{2}{100} = \frac{m}{m+55.5}$

m-50m= -55.5

49m=55.5

m = 1.13

Answer 2

Explanation:

this is the correct answer